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Maximum stress that can be applied to a ...

Maximum stress that can be applied to a wire which supports on elevator is `sigma`. Mass of elevator is `m ` and it is moved upwards with an acceleration of `g//2`. Minimum diameter of wire ( Neglecting weight of wire ) must be

A

`sqrt(2mg)/(pisigma)`

B

`sqrt(3mg)/(2pisigma)`

C

`sqrt(5mg)/(2pisigma)`

D

`sqrt((6mg)/(pisigma))`

Text Solution

Verified by Experts

The correct Answer is:
D
Force on wire = Load carried
`= m(g+a) = m(g+g/2) = (3 mg)/2`
Stress `sigma = F/A rArr A = F/sigma`
Hence, `A_(min) = F/sigma_(max)` ltb rgt ` (pi d_(min)^(2))/(4) = (3/2 m.g)/sigma`
`rArr d_(min)^(2) = (3/2 mgxx4)/(sigmapi)`
` d_(min) = sqrt(6/pi. (mg)/(sigma)`
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